I know this might be very stupid question but I have been spending hours on this
want to read a .csv file that I dont have its full path (*/*data.csv). I know that following would get the path of the current directory but don't know how to adapt
Marks <- read.csv(dir(path = '.', full.names=T, pattern='^data.*\\.csv'))
tried this one as well but not working
Marks <- read.csv(file = "*/*/data.csv", sep = ",", header=FALSE))
I can't identify specific path as this will be used on different machines with different paths but I am sure about the sub-folders of the main directory as they are result of a bash script
and I am planing to call this from within unix which defines the workspace
my data structure is
lecture01/test/data.csv
lecture02/test/data.csv
lecture03/test/data.csv
Your comments -- though not currently your question itself -- indicate you expect to run your code in a working directory that contains some number of subdirectories (lecture01, lecture02, etc), each of which contain a subdirectory 'marks' that in turn contains a data.csv file. If this is so, and your objective is to read the csv from within each subdirectory, then you have a couple of options depending on the remaining details.
Case 1: Specify the top-level directory names directly, if you know them all and they are potentially idiosyncratic:
dirs <- c("lecture01", "lecture02", "some_other_dir")
paths <- file.path(dirs, "marks/data.csv")
Case 2: Construct the top-level directory names, e.g. if they all start with "lecture", followed by a two digit number, and you are able to (or specifically wish to) specify a numeric range, e.g. 01 though 15:
dirs <- sprintf("lecture%02s", 1:15)
paths <- file.path(dirs, "marks/data.csv")
Case 3: Determine the top-level directory names by matching a pattern, e.g. if you want to read data from within every directory starting with the string "lecture":
matched.names <- list.files(".", pattern="^lecture")
dirs <- matched.names[file.info(matched.names)$isdir]
paths <- file.path(dirs, "marks/data.csv")
Once you have a vector of the paths, I'd probably use lapply to read the data into a list for further processing, naming each one with the base directory name:
csv.data <- lapply(paths, read.csv)
names(csv.data) <- dirs
Alternatively, if whatever processing you do on each individual CSV is done just for its side effects, such as modifying the data and writing out a new version, and especially if you don't ever want all of them to be in memory at the same time, then use a loop.
If this answer misses the mark, of even if it doesn't, it would be great if you could clarify the question accordingly.
I have no code but I would do a reclusive glob from the root and do a preg_match to find the .csv file (use glob brace).
Related
I have a task that requires me to use a specific column in a CSV spreadsheet that stores the file names, for example:
File Name
CA-001
WV-001
ma-001
My task is to move some files from folder 'source' to folder 'target'.
And I'm using this csv spreadsheet as a crosswalk to select any files with names that match with what's in the column 'File Name'. Then I'm asking R to copy from the source folder that contains not only these files but also other files that are not in this list(eg: CO-001, SC-001...). If it's helpful, all of the files are PDFs, so we don't worry about file type. I want only the files that have names match with what's in the csv spreadsheet. How can I do this?
I have some sample code below, but it still didn't execute successfully.
source <- "C:/Users/53038/MovePDF/Test_From"
target <- "C:/Users/53038/MovePDF/Test_To"
all.files <- list.files(path = source)
csvfile <- read.csv('C:/Users/53038/MovePDF/Master.csv')
toCopy <- all.files[all.files %in% csvfile$Move]
file.copy(toCopy, target)
Thank you!
With the provided code, the selection of patterns you want to match will be in csvfile$File.Name.
I'm assuming the source directory is potentially very large. Instead of performing slow regular expressions to match substrings (while we know the exact filename), and/or getting a complete file listing (which is also slow), I will only seek if the exactly wanted filenames exist before copying them:
source <- "C:/Users/53038/MovePDF/Test_From"
target <- "C:/Users/53038/MovePDF/Test_To"
csvfile <- read.csv('C:/Users/53038/MovePDF/Master.csv')
# add .pdf suffix
toCopy <- paste0(csvfile$File.Name,'.pdf')
# add source directory path
toCopy <- file.path(source, toCopy)
# optional: extract only the existing files from toCopy. You can skip this step if you're sure they exist and/or you don't mind receiving errors
toCopy <- toCopy[file.exists(toCopy)]
# make it so
file.copy(toCopy, target, overwrite = T)
I would preferably keep the .pdf extension in the filename at all times, so also in the source CSV. There would be an issue on case-sensitive filesystems (almost all Linux installations, rarely macOS or Windows) if the extension is .PDF, .Pdf, etc.
Need to write a piece of code in R that'll create a list specifying:
Names of subfolders with a pre-set depth (e.g. 2 levels down)
Path
Date modified
I've tried to use the following generic function but had no luck:
list.files(path, pattern=NULL, all.files=FALSE,
full.names=FALSE)
dir(path, pattern=NULL, all.files=FALSE,
full.names=FALSE)
Would very much appreciate your response.
I think what your are missing is the recursive = TRUE parameter in list.files()
One possible solution could be to list all files first and then limit the output to 2 levels accordingly.
files <- list.files(path = "D:/cmder/", recursive = TRUE)
Since R represents paths using "/" a simple example could be to remove everything that has more than 3 slashes if you need a depth of 2.
files[!grepl(".*/.*/.*/.*", files)]
Be careful on windows as you might see a back slash "\" there sometimes, only if your path information comes from something different that R itself, e.g. a csv import.
My grepl() statement can probably be improved as I'm not an expert there.
I'm writing a loop script which involves reading a file from a workbook (using the package XLConnect). The challenge is that the file names contain characters (representing time) that I want to ignore.
For example, here are 3 paths to those files:
G://User//Documents//daily_data//Op_Schedule_20160520_132025.xlsx
G://User//Documents//daily_data//Op_Schedule_20160521_142805.xlsx
G://User//Documents//daily_data//Op_Schedule_20160522_103052.xlsx
I need to import hundreds of those files. I can easily account for the character string representing the date (e.g. 20160522), but not the time.
Is there a way to tell R to ignore some characters located in the file path? Here is how I was thinking of writing my script (the "???" is where i need help). I know a loop is probably not the most efficient way, but i'm open to suggestions, should you have any:
require(XLConnect)
path= "G://User//Documents//daily_data//Op_Schedule_"
wd.seq = format(seq(as.Date("2014-01-01"),as.Date("2016-12-31"),"days"),format="%Y%m%d")
scheduleList = rep(list(matrix(1,1,1)),length(wd.seq))
for(i in 1:length(wd.seq)) {
wb = loadWorkbook(file= paste0(path,wd.seq[i],"???",".xlxs"))
scheduleList[[i]] = readWorksheet(wb,sheet='=SCHEDULE', header = TRUE)
}
`
Thanks for reading and suggestions, if any.
Mathieu
I don't know if this is helpful, but if you want to read all the files in a certain directory (which it seems to me is what you're after), you can read all the filenames into a list using the list.files() function, for example
fileList <- list.files(""G://User//Documents//daily_data//")
And then load the xlsx files looping through the list with a for loop
for(i in fileList) {
loadWorkbook(file = i)
}
I haven't used the XLConnect function before so that exact code probably doesn't work, but the loop will iterate through all the files in that directory and so you can construct your loading call using the i variable for the filename (it won't be an absolute path though, so you might need to use paste to add the first part of the filepath)
I realize there might be other files in the directory that are not excel files, you could use grepl to select only files containg "OP_Schedule_"
fileListClean <- fileList[grepl("Op_Schedule_",fileList)]
or perhaps only selecting .xlsx files in the directory:
fileListClean <- fileList[grepl(".xlsx",fileList)]
Edit to fit your reply:
Since you need to fit it to a sequence, you can do it as you did earlier:
wd.seq = format(seq(as.Date("2014-01-01"),as.Date("2016-12-31"),"days"),format="%Y%m%d")
wd.seq2 <- paste("Op_Schedule_", wd.seq, sep = "")
And then use grepl to only pick files starting with that extensions:
fileListClean <- fileList[grepl(paste(wd.seq2, collapse = "|"), fileList)]
Full disclosure: The last part i got from this SO answer: grep using a character vector with multiple patterns
Using this script I have created a specific folder for each csv file and then saved all my further analysis results in this folder. The name of the folder and csv file are same. The csv files are stored in the main/master directory.
Now, I have created a csv file in each of these folders which contains a list of all the fitted values.
I would now like to do the following:
Set the working directory to the particular filename
Read fitted values file
Add a row/column stating the name of the site/ unique ID
Add it to the masterfile which is stored in the main directory with a title specifying site name/filename. It can be stacked by rows or by columns it doesn't really matter.
Come to the main directory to pick the next file
Repeat the loop
Using the merge(), rbind(), cbind() combines all the data under one column name. I want to keep all the sites separate for comparison at a later on stage.
This is what I'm using at the moment and I'm lost on how to proceed further.
setwd( "path") # main directory
path <-"path" # need this for convenience while switching back to main directory
# import all files and create a character type array
files <- list.files(path=path, pattern="*.csv")
for(i in seq(1, length(files), by = 1)){
fileName <- read.csv(files[i]) # repeat to set the required working directory
base <- strsplit(files[i], ".csv")[[1]] # getting the filename
setwd(file.path(path, base)) # setting the working directory to the same filename
master <- read.csv(paste(base,"_fiited_values curve.csv"))
# read the fitted value csv file for the site and store it in a list
}
I want to construct a for loop to make one master file with the files in different directories. I do not want to merge all under one column name.
For example, If I have 50 similar csv files and each had two columns of data, I would like to have one csv file which accommodates all of it; but in its original format rather than appending to the existing row/column. So then I will have 100 columns of data.
Please tell me what further information can I provide?
for reading a group of files, from a number of different directories, with pathnames patha pathb pathc:
paths = c('patha','pathb','pathc')
files = unlist(sapply(paths, function(path) list.files(path,pattern = "*.csv", full.names = TRUE)))
listContainingAllFiles = lapply(files, read.csv)
If you want to be really quick about it, you can grab fread from data.table:
library(data.table)
listContainingAllFiles = lapply(files, fread)
Either way this will give you a list of all objects, kept separate. If you want to join them together vertically/horizontally, then:
do.call(rbind, listContainingAllFiles)
do.call(cbind, listContainingAllFiles)
EDIT: NOTE, the latter makes no sense unless your rows actually mean something when they're corresponding. It makes far more sense to just create a field tracking what location the data is from.
if you want to include the names of the files as the method of determining sample location (I don't see where you're getting this info from in your example), then you want to do this as you read in the files, so:
listContainingAllFiles = lapply(files,
function(file) data.frame(filename = file,
read.csv(file)))
then later you can split that column to get your details (Assuming of course you have a standard naming convention)
I am stuck. I need a way to iterate through a bunch of subfolders in a directory, pull out 4 .csv files , bind the contents of those 4 .csv files, then write out the new .csv to a new directory using the name of the initial subfolder as the name of the new .csv.
I know R could do this. But I am stuck at how to iterate across the subfolders and bind the csv files together. My obstacle is that each subfolder contains the same 4 .csv files using the same 8-digit id. For example, subfolder A contains 09061234.csv, 09061345.csv, 09061456.csv, and 09061560.csv. subfolder B contains 9061234.csv, 09061345.csv, 09061456.csv, and 09061560.csv. (...). There are 42 subfolders, and hence 168 csv files with the same names. I want to compact the files down to 42.
I can use list.files to retrieve all the subfolders. But then what?
##Get Files from directory
TF = "H:/working/TC/TMS/Counts/June09"
##List Sub folders
SF <- list.files(TF)
##List of File names inside folders
FN <- list.files(SF)
#Returns list of 168 filenames
###?????###
#How to iterate through each subfolder, read each 8-digit integer id file,
#bind them all together into one single csv,
#Then write to new directory using
#the name of the subfolder as the name of the new csv?
There is probably a way to do this easily but I am a noob with R. Something involving functions, paste and write.table perhaps? Any hints/help/suggestions is greatly appreciated. Thanks!
You can use recursive=T option for list.files,
lapply(c('1234' ,'1345','1456','1560'),function(x){
sources.files <- list.files(path=TF,
recursive=T,
pattern=paste('*09061*',x,'*.csv',sep='')
,full.names=T)
## ou read all files with the id and bind them
dat <- do.call(rbind,lapply(sources.files,read.csv))
### write the file for the
write(dat,paste('agg',x,'.csv',sep='')
}
After some tweaking of agstudy's code, I came up with the solution I was ultimately after. There were a couple of missing pieces that are more due to the nature of my specific problem, so I am leaving agstudy's answer as "accepted".
Turns out a function really wasn't needed. At least not for now. If I need to perform this same task again, I will create a function out of it. For now, I can solve this particular problem without it.
Also, for my instance, I needed a conditional "if" statement to handle any non-csv files that may have lived in the subfolders. By adding an if statement, R throws warnings and skips any files that are not comma-separated.
Code:
##Define directory path##
TF = "H:/working/TC/TMS/Counts/June09"
##List of subfolder files where file name starts with "0906"##
SF <- list.files(TF,recursive=T, pattern=paste("*09061*",x,'*.csv',sep=""))
##Define the list of files to search for##
x <- (c('1234' ,'1345','1456','1560')
##Create a conditional to skip over the non-csv files in each folder##
if (is.integer(x)){
sources.files <- list.files(TF, recursive=T,full.names=T)}
dat <- do.call(rbind,lapply(sources.files,read.csv))
#the warnings thrown are ok--these are generated due to the fact that some of the folders contain .xls files
write.table(dat,file="H:/working/TC/TMS/June09Output/June09Batched.csv",row.names=FALSE,sep=",")